Question 216: Magnetism and Matter

Question

The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the
loop is B1. When the dipole moment is doubled by keeping the current constant, the magnetic field at the
centre of the loop is $${{B_2}}$$. The ratio $${{{B_1}} \over {{B_2}}}$$ is:

Accepted Answer

Dipole moment, M = IA
Let radius of circular loop = R
$$	herefore\,\,\,$$ M  =  I $$ 	imes $$ $$\pi $$R
2
Later,  we keep current constant ,
But dipole moment becomes double, let new radius = R
1
$$	herefore\,\,\,$$ 2M  =  I $$ 	imes $$ $$\pi $$R$$_1^2$$
$$ \Rightarrow $$$$\,\,\,$$ 2I$$\pi $$R
2
=  I$$\pi $$R$$_1^2$$
$$ \Rightarrow $$$$\,\,\,$$ R
1
=  $$\sqrt 2 $$ R
At the center of circular ring, the magnetic field,
B  =  $${{{\mu _0}I} \over {2R}}$$
$$	herefore\,\,\,$$ B
1
=  $${{{\mu _0}I} \over {2R}}$$   and    B
2
= $${{{\mu _0}I} \over {2 	imes \left( {\sqrt 2 R} ight)}}$$
$$	herefore\,\,\,$$ $${{{B_1}} \over {{B_2}}}$$  =   $${{{{{\mu _0}I} \over {2R}}} \over {{{{\mu _0}I} \over {2\sqrt 2 \,R}}}}$$
= $${{2\sqrt 2 } \over 2}$$
= $$  \sqrt 2 $$

Answer

2

Answer

$$\sqrt 3 $$

Answer

$$\sqrt 2 $$

Answer

$$1 \over \sqrt 2 $$

Magnetism and Matter

The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is B1. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is ${{B_2}}. The ratio {{{B_1}} \over {{B_2}}}$ is:

Choose the correct answer: